# Rearranging Differential Equation

• Jan 18th 2011, 03:28 PM
Miika89
Rearranging Differential Equation
Hi, I was going over my notes and didnt understand one of the steps made in a calculation, was hoping someone could explain it to me.

Im trying to find an analytic solution for an ODE: dY/dX + X*Y = X
The integrating factor is I = e^((X^2)/2)

Multiplying the equation with the integrating factor

I*(dY/dX) + X*I*Y = X*I

This next step I dont follow where it gets simplified to:

(d/dX)*(Y*e^((X^2)/2)) = X*(e^((X^2)/2))

The following line's after are:

YI = I

Y = 1 + C*I

Any help would be appriciated, thank you.
• Jan 18th 2011, 03:49 PM
Ackbeet
The step you're wondering about is the heart of the integrating factor procedure. Suppose you had the DE

$\dfrac{dy}{dx}+P(x)\,y=Q(x),$ and you multiply through by the integrating factor $e^{\int P(x)\,dx}.$ Then you get

$e^{\int P(x)\,dx}\dfrac{dy}{dx}+e^{\int P(x)\,dx}P(x)\,y=Q(x)e^{\int P(x)\,dx}.$

The whole point of the procedure is that the LHS is now a total derivative. Indeed, if we examine

$\dfrac{d}{dx}\left[e^{\int P(x)\,dx} y\right]=\left(e^{\int P(x)\,dx}\right)\left(\dfrac{dy}{dx}\right)+y\left (\dfrac{d}{dx}e^{\int P(x)\,dx}\right)=
e^{\int P(x)\,dx}\,\dfrac{dy}{dx}+ye^{\int P(x)\,dx}\,P(x),$

which is just our new LHS after we multiplied through by the integrating factor.

Someone was clever enough to discover that the integrating factor was indeed $e^{\int P(x)\,dx},$ and after that, it became a procedure.

Does that clear things up a bit for you, perhaps?
• Jan 18th 2011, 03:55 PM
pickslides
Quote:

Originally Posted by Miika89
This next step I dont follow where it gets simplified to:

(d/dX)*(Y*e^((X^2)/2)) = X*(e^((X^2)/2))

If you consider the product rule before this line you had

$vu'+uv' = f(x)$

$(uv)' = f(x)$

And integrating both sides

$uv = \int f(x)$
• Jan 18th 2011, 03:56 PM
Miika89
Yes, thank you.

If I didnt know this procedure is there a way of getting from I*(dY/dX) + X*I*Y = X*I to YI = I? Or would you suggest its too complicated and I just memorise how this procedure works?
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Thank you both, I think pickslides just answered my 2nd question
• Jan 18th 2011, 03:58 PM
pickslides
Quote:

Originally Posted by Miika89
Yes, thank you.

If I didnt know this procedure is there a way of getting from I*(dY/dX) + X*I*Y = X*I to YI = I? Or would you suggest its too complicated and I just memorise how this procedure works?

I say memorise it. It comes in very handy and enhances your understanding of calculus.
• Jan 18th 2011, 04:17 PM
Prove It
I'd say as well as memorising it i.e. that after you have multiplied by the Integrating Factor the LHS reduces to $\displaystyle \frac{d}{dx}(I\,y)$, get good at recognising Product Rule expansions, i.e. that $\displaystyle u\,\frac{dv}{dx} + v\,\frac{du}{dx} = \frac{d}{dx}(u\,v)$.
• Jan 19th 2011, 04:01 AM
Miika89
Not having gotten much further, Im stuck on how this next stage works:

y*I = I
y = 1 + C*e^-((x^2)/2)

I am assuming he has now integrated the RHS?
• Jan 19th 2011, 12:12 PM
pickslides
Given $\displaystyle I= e^{\frac{x^2}{2}}$

$\displaystyle Iy'+Ixy = xI$

Note $\displaystyle I'=Ix$ then

$\displaystyle (Iy)' = xI$

$\displaystyle Iy = \int xI~dx$

$\displaystyle e^{\frac{x^2}{2}}y = \int xe^{\frac{x^2}{2}}~dx$

Then by substitution $\displaystyle u = \frac{x^2}{2}$ integrate the right hand side.

Final step is to divide both sides by $\displaystyle e^{\frac{x^2}{2}}$